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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions

2 votes
Accepted

Trying to solve for the remainder of $a^q$ modulo $q$

As I pointed in the comments, the equation in question for $a\not\equiv0\pmod{q}$ can be restated as $$a + 2af_q(a) \equiv c\pmod{q},$$ where $f_q(a)$ is the Fermat quotient of $a$ with respect to $q$ …
Max Alekseyev's user avatar
14 votes
Accepted

When is $\mathrm{gcd}(k,p^k-1)=1$ true?

It is easier to describe non-good (bad) numbers with respect to a given prime $p$. For each such number $k$, there exists a prime $q$ such that $q\mid k$ and $q\mid (p^k - 1)$. It follows that $k$ is …
Max Alekseyev's user avatar
1 vote
Accepted

Summing the max value of the distinct pairs in a multiset

This is a corrected and extended version of my earlier comment. Let $A = \{ b_1^{m_1}, \dots, b_k^{m_k}\}$ where $b_1 < \dots < b_k$ and $m_i$ are their multiplicities with $\sum_{i=1}^k m_i = n$. The …
Max Alekseyev's user avatar
10 votes
Accepted

Ordinary partitions vs partitions into odd parts

The g.f. for the right-hand is $$e^{2x}\cdot e^{2x^2} \cdots = e^{2\frac{x}{1-x}}.$$ For the left-hand side, additionally introducing variable $y$ to account for partition length, we get $$\sum_{n\ge …
Max Alekseyev's user avatar
0 votes

How much do these interval collections cover?

Equivalently we can define a covered number $n$ as such that there exists a prime $p<\sqrt{n}$ such that $p\nmid n$ and $\lceil n/p\rceil$ is also prime. Indeed, taking $q:=\lceil n/p\rceil$ we have t …
Max Alekseyev's user avatar
4 votes

Are these equations solvable in positive integers?

For (a), we can introduce $s:=x+1$ and pose the question as $(sy-1)\mid (s^4 - 3s^3 + 3s^2 - 3s + 1)$. Equivalently, we have that $$\frac{s^2 + y^2 - 3s - 3y + 3}{sy-1}$$ is an integer. This problem i …
Max Alekseyev's user avatar
13 votes
Accepted

When is the number of areas obtained by cutting a circle with $n$ chords a power of $2$?

Denoting $k:=2^{\lfloor m/2\rfloor}$, we get two cases to consider: $k^2 = f(n)$ and $2k^2 = f(n)$, or making the coefficients integer: $$(12k)^2 = 144f(n)\qquad\text{and}\qquad (12k)^2 = 72f(n).$$ Th …
Max Alekseyev's user avatar
3 votes
Accepted

Special configurations on a circle from a homological algebra problem

There is a simple characterization of interesting configurations: Lemma. A configuration $x_0=0< x_1 < x_2 < ... <x_r$ of Gorenstein dimension $g$ is interesting if and only if there exist indices $i, …
Max Alekseyev's user avatar
2 votes

Calculate the great common factor between $2^{2n+1}-1$ and $2^{4m+2}+1$

In general we have $$\gcd(2^a + 1, 2^b - 1) = \frac{2^{\gcd(2a,b)}-1}{2^{\gcd(a,b)}-1},$$ which is evaluates to $1$ for $a=4m+2$ and $b=2n+1$.
Max Alekseyev's user avatar
9 votes

Does $a_0=6$, $a_{n+1} = (a_n-1)\cdot a_n\cdot (a_n+1)$ define a square-free sequence?

The question is equivalent to asking whether $b(n):=\frac{a(n)}{a(n-1)}$ are squarefree. The sequence $b(n)$ is listed in OEIS A231831 and satisfies the recurrence $b(n+1) = b(n)^3 + b(n)^2 - 1$ with …
Max Alekseyev's user avatar
2 votes
Accepted

The existence of solutions of a system of indeterminate equations

In the comments OP proposed a greedy algorithm to represent a given positive integer $A$ as the sum of triangular numbers whose indices sum to $m$, and applied it to $A = \frac{m(m+1)}4$. I will prove …
Max Alekseyev's user avatar
1 vote

The existence of solutions of a system of indeterminate equations

I expect that in many cases the following construction will do the job. By Fermat polygonal number theorem, we have representation of $\frac{m(m-3)}4$ as the sum of 3 triangular numbers: $$ \frac{m(m- …
Max Alekseyev's user avatar
30 votes

Proof for new deterministic primality test

The claim does not hold. A counterexample is given by $n=14$, $p=134123250258009499$ and correspondingly $$N = 2197475332227227631617 = 193 \cdot 12289 \cdot 926510094425921.$$ It can be easily verifi …
Max Alekseyev's user avatar
4 votes
Accepted

What work can be done to study the solutions of $\varphi\left(x^{\sigma(x)}\sigma(x)^x\right...

Here is a proof that if an odd integer $x>1$ satisfies (1), then $x$ is a perfect number. First, by using the property that $\varphi(nm)=n\varphi(m)$ whenever $\mathrm{rad}(n)\mid\mathrm{rad}(m)$, we …
LSpice's user avatar
  • 10.6k
2 votes
Accepted

Coefficients of number of the same terms which are arising from iterations based on binary e...

In other words, if $(b_\ell b_{\ell-1}\dots b_0)_2$ is the binary representation of $n$, then $$a(n) = g(g(\dots g(g(0,b_0),b_1)\dots ),b_{\ell-1}), b_\ell),$$ where $$g(A,b) = \begin{cases} A+2, &\te …
Max Alekseyev's user avatar
1 vote

Solutions of a linear diophantine equation

$N(h)$ can be expressed via partition function $q$ as $$N(h)=q(6h-6,h-1).$$
Max Alekseyev's user avatar
1 vote

Algorithm for finding integers in a range with multiples in a short interval

Depending on the value of $D$ as compared to $X$, you may want to switch from scanning the interval $[D+1,2D]$ to the interval of co-factors: $\big[ \lceil \frac{X}{2D} \rceil, \lfloor \frac{X+H}{D+1} …
Max Alekseyev's user avatar
3 votes
Accepted

Lucas–Lehmer test and triangle of coefficients of Chebyshev's

By the composition property of Chebyshev polynomials $T_m(T_n(x))=T_{mn}(x)$. Since $x^2-2 = 2T_2(\tfrac{x}2)$, we have $S_i = 2T_{2^i}(2)$ for all $i\geq 0$. Furthermore, since $T_{2^k}(x) = \frac{U_ …
Max Alekseyev's user avatar
4 votes

Non-Wieferich primes with Euler quotient modulo $p$ two and alternating harmonic numbers

While no composite terms of A128465 are known, here is a proof that an odd prime $p$ belongs to A128465 if and only if $b(p)\equiv 2(-1)^{\tfrac{p+1}2}\pmod{p}$. First notice that for an odd prime $p$ …
Max Alekseyev's user avatar
1 vote
Accepted

Asymptotic for a sum involving GCD and Euler totient function

Let $c:=\gcd(\phi(d),d)$. then setting $r:=dk$ the sum can be rewritten as $$c\sum_{k\leq x/d} \gcd(\tfrac{\phi(d)}c,k) \approx \frac{c^2x}{d\phi(d)}f(\tfrac{\phi(d)}c),$$ where $f(m) := \sum_{k=1}^m …
Max Alekseyev's user avatar
3 votes

How many ways to pick k integers with fixed sum and product

Let $B_k({\cal S},{\cal P},{\cal X})$ be a bound for the number of solutions for any $S\leq {\cal S}$, $P\leq {\cal P}$ and ${\cal X} \leq x_1 \leq x_2 \leq \dots \leq x_k$. Using Iverson's bracket no …
Max Alekseyev's user avatar
3 votes
Accepted

Counting numerical semigroups by largest element of minimal generating set

A key observation is that two sets of generators $g, g'\subseteq [n]$ produce the same semigroup if and only if $\langle g\rangle \cap [n] = \langle g'\rangle \cap [n]$. Hence, the number of different …
Max Alekseyev's user avatar
3 votes
Accepted

A query about modular arithmetic on a matrix

The question is equivalent to finding an integer vector $x$ such that $$xM = \iota_K,$$ where $\iota_k$ is the all-1 vector of length $K$. By Rouché–Capelli theorem, this equation has a solution modul …
Max Alekseyev's user avatar
4 votes
Accepted

Expanding in Fibonacci powers

Here is the direct proof without using the result of Ardila. We will represent a sum of distinct Fibonacci numbers $m=\sum_{j=2}^n b_{j-1} F_j$, where $b_j\in\{0,1\}$, as a bit string $b_1b_2\ldots b_ …
Max Alekseyev's user avatar
2 votes
Accepted

Partition of $(2^{n+1}+1)2^{2^{n-1}+n-1}-1$ into parts with binary weight equals $2^{n-1}+n$

Notice that for $i\in\{0,1,\dots,2^{n-1}+n\}$ we have $$a(i+1,2^{n-1}+n) = 2^{2^{n-1}+n+1} - 1 - 2^{2^{n-1}+n-i}.$$ Then the sum in question can be easily computed: \begin{split} & a(1,2^{n-1}+n)+\sum …
Max Alekseyev's user avatar
2 votes

Existence of solution for a system of quadratic diophantine equations / symmetric quadratic ...

UPDATE. Using factorization $-2(x+1)^2x^{p-3}$ over the corresponding number field, I established that there are no solutions for $p=17$. Furthermore, I computationally verified that for primes $p<30$ …
Max Alekseyev's user avatar
3 votes

Only trivial solutions to system of linear diophantine equations possibly related to hamilto...

Below I show that $m$ cannot be constant. The given system of $m$ equations can be reduced to the case of $m=1$, that is, to a single equation: $$\sum_{j=1}^n y_j a'_{j} = \sum_{j=1}^n a'_{j}$$ where …
Max Alekseyev's user avatar
3 votes
Accepted

A diophantine equation inspired in a conjecture due to Gica and Luca, example of a large Mer...

The equation (1) for a fixed $x$ is equivalent to the congruence: $$y^2 \equiv 2(1+z^2)\pmod{x}.$$ For $x=25964951$, we have $2\equiv 3328351^2\pmod{x}$, and thus all solutions are obtained from those …
Max Alekseyev's user avatar
7 votes
Accepted

Small covering of divisors

UPDATED We can simply take $$B = \{1\} \cup \{ d\in D_n\ :\ d > n^{1/2} \}.$$ Then for any $d\in D_n$: if $d\leq n^{1/2}$, we take $(a,b)=(d,1)$; if $d>n^{1/2}$, we take $(a,b)=(1,d)$. Then $$\sum_{ …
Max Alekseyev's user avatar
10 votes

Primality test for numbers of the form $\frac{a^p-1}{a-1}$?

The test fails already for $p=3$ and $a=7$, claiming that $57=3\cdot 19$ is prime. ADDED. And new test fails for $a=10$ and $p=5$. ADDED#2. And the test in EDIT 2 fails for $a=52$ and $p=3$.
Max Alekseyev's user avatar
5 votes

Primality test for $\frac{(10 \cdot 2^n)^m-1}{10 \cdot 2^n-1} - 2$ and $\frac{(10 \cdot 2^n)...

It's often the case with such tests that the "only if" part is more or less easy to prove, while the "if" part is inaccessible for proving or disproving. Below I prove the "only if" part, ie. assuming …
Max Alekseyev's user avatar
6 votes

Resources where I can find open problems in number theory along with their level of difficulty

Some open problem collections from my bookmarks: Number Theory @ Open Problem Garden List of unsolved problems in number theory @ Wikipedia Unsolved problems @ MathWorld Unsolved Problems and Rewards …
Max Alekseyev's user avatar
5 votes

Parametrization of integral solutions of $3x^2+3y^2+z^2=t^2$ and rational solutions of $3a^2...

For #1, we can take a particular solution such as $(a_0,b_0,c_0)=(0,0,1)$, and search parametric solution in the form: $(a,b,c)=(a_0+\alpha t, b_0+\beta t, c_0+t)$. Plugging it into the equation and s …
Max Alekseyev's user avatar
5 votes
Accepted

Sequences that sums up to second differences of Bell and Catalan numbers

Let me address the case of $s_2(n)$. First we notice that $g(n-1) = k$ whenever $n=(2k+1)2^t$. Then for $n=2^{t_1}(1+2^{1+t_2}(1+\dots(1+2^{1+t_\ell}))\dots)>1$ with $t_j\geq 0$, we have $$a_2(n) = \b …
Max Alekseyev's user avatar
3 votes
Accepted

Constructing an integer with small residues for two distinct primes in polynomial time

If such $m$ exists, then $m=up+a=vq+b$ for some $u,v\in O(T^{1/2+\epsilon})$ and $a,b\in O(\mathrm{polylog}(T))$. Then $up-vq=b-a$ and thus $\frac{p}q - \frac{v}u=\frac{b-a}{uq}$, implying that $\frac …
Max Alekseyev's user avatar
6 votes
Accepted

On roots of irreducible quadratics modulo composites

Ability to find all roots leads to finding factorization of $N$. For example, if $N=pq$ is a product of two odd primes, and we find a root $x'$ of $x^2-1\equiv 0\pmod N$ such that $x'\equiv 1\pmod p$ …
Max Alekseyev's user avatar
0 votes
Accepted

Is there a general way to solve this modular equation?

There seems to be no simple formula for the solutions to the given congruence. Still, they can be computed iteratively as follows. First, we notice that $2^m = (3-1)^m \equiv (-1)^m \sum_{i=0}^{N-1} \ …
Max Alekseyev's user avatar
1 vote

A variant of Landau's function

Let $k = \nu_2(g(n))$. Then $g(n)$ is attained at $n = 2^k + \texttt{1s and powers of odd primes}$, implying that $g(n) = h(n-2^k)2^k$. Hence, $h(n) = \frac{g(n+2^k)}{2^k}$ for some $k$ satisfying $k= …
Max Alekseyev's user avatar
3 votes

Complexity of a Fibonacci numbers discrete log variation

The Binet formula for Fibonacci numbers is $$F_n = \frac{\phi^n - (-\phi)^{-n}}{\phi - (-\phi)^{-1}},\qquad\text{where}\ \phi:=\frac{1+\sqrt{5}}2.$$ Then the congruence $F_n\equiv k\pmod{m}$ reduces t …
Max Alekseyev's user avatar
5 votes
Accepted

Number of positive integers $k$ such that there exists a nonnegative integer $m$ with $k + k...

The formula $c(n)=a(n+1)$ is pretty much straightforward, noticing that $$\lfloor \log_{i+1}(n-i)\rfloor - \lfloor\log_{i+1}(n-1-i)\rfloor=1\quad\text{iff}\quad n-i=(i+1)^m\text{ for some }m.$$ The l …
Max Alekseyev's user avatar
4 votes
Accepted

direct proof of an identity regarding certain symmetry of integer partitions?

I assume you meant $j_t\leq a_t$ (not $j_t<a_t$). It's sufficient to prove that for any analytic function $f(x)$, the function $$F(a_1,a_2) := \sum_{l\geq 0} \sum_{j=0}^{a_2} \frac{(-1)^ll!}{(a_1+l+1) …
Max Alekseyev's user avatar
9 votes

$23005\cdot (2^n-1)\cdot 2^n +1=p^2$

Multiplying the equation by $2^n$ and denoting $X:=2^n$ and $Y:=p2^{\lfloor n/2\rfloor}$, we obtain two elliptic curves (depending on the parity of $n$): $$(23005(X-1)X+1)X=Y^2,$$ $$(23005(X-1)X+1)X=2 …
Max Alekseyev's user avatar
3 votes
Accepted

Least modulus distinguishing some integers

I'm not sure about the state of art, but here is a rough estimate in terms of $c$ and $\ell:=a_c-a_1$ in the case $c\ll \ell$. First, we notice that for an integer $M$ satisfying $$M\# ~>~ \big(\frac{ …
Max Alekseyev's user avatar
3 votes
Accepted

Runs of consecutive numbers all of which are Murthy numbers

Let $m+1, \dots, m+n$ be a sequence of $n$ consecutive Murthy numbers such that each $m+i$ shares with its reversal $\overline{m+i}$ a prime factor $p_i\equiv 3\pmod4$ such that 10 is a quadratic nonr …
Max Alekseyev's user avatar
3 votes

What is the highest power of 2 that divides $3^y(2z-1)-1$?

I doubt there is a simple expression for $x$ as a function of $y,z$. However, it is possible to characterize all cases when $2^k\mid 3^y (2z-1) - 1$ for a given positive integer $k$. Say, for $k\geq 3 …
Max Alekseyev's user avatar
23 votes
Accepted

Are there infinitely many positive integer solutions to $(3+3k+l)^2=m\,(k\,l-k^3-1)$?

It does have infinitely many positive solutions. Here is just one such series. Consider the following recurrence sequence: $$u_0=1,\ u_1=2,\ u_{n+1} = 23 u_n - u_{n-1} - 4\qquad (n\geq 1).$$ Let $t,k$ …
Max Alekseyev's user avatar
6 votes

Convergence of $\sum(n^p\sin^qn)^{-1}$

A while ago I've addressed this question in On convergence of the Flint Hills series.
Max Alekseyev's user avatar
5 votes
Accepted

Analogue of Fermat's little theorem for Bernoulli numbers

A proof is essentially given in Section 5.1 of Notes on primitive lambda-roots by P. J. Cameron and D. A. Preece.
Max Alekseyev's user avatar
7 votes

For which $n$ is $\sum_{k=1}^n 1 / \varphi(k)$ an integer?

Here is some computational evidence that $n\in\{1,2,4\}$ are the only $n$ that deliver an integer sum. For an odd prime $p$, let $a_p < b_p$ denote two smallest even positive integers such that $a_pp+ …
Max Alekseyev's user avatar
24 votes

When do binomial coefficients sum to a power of 2?

I doubt this problem has an easy solution. It is clear how it was approached for small fixed $N$. Below I show how it can be addressed for the case of fixed odd $n>1$. When $n>1$ is odd, $S(N,n)$ as a …
Max Alekseyev's user avatar

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